Uranium metal can be produced by reaction of uranium tetrafluoride (UF4) with calcium metal (Ca). The by-product of this reaction is calcium fluoride (CaF2). The reactor is charged with 292.0 kg of UF4 and 47.0 kg of calcium. Assume that the reaction goes to completion.
The reactor contents at the completion of this reaction will be the following:
Enter the number of kg of UF4 remaining:
- Enter the number of kg of Ca remaining:
- Enter the number of kg of U present:
- Enter the number of kg of CaF2 present:
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. This is a standard LR problem. What do you not understand?
If is limit what a reaction with reagent and excess but don't know how to start this exercise, please help me
UF4 + 2Ca ==> 2CaF2 + U
mols UF4 initially = grams/molar mass = 292,000/314 = approx 930 but you need to do this more accurately. All of the following calculations are estimates; recalculate those also.
mols Ca = grams/atomic mass = 47,000/40.1 = estimated 1180
Using the coefficients in the balanced equation, convert mols UF4 to mols U. That's 930 x (1 mol U/1 mol UF4) = estimated 930 mols U.
Do the same for converting mols Ca to mols U. That's 1180 x (1 mol U/2 mols Ca) = 1180 x 1/2 = estimated 590 mols U.
You can see that mols U don't agree; therefore, one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR. Therefore, Ca is the limiting reagent and estimated 590 mols U will be formed.
g U formed = mols x atomic mass = 590 x 238 = estimated 140,000 grams U.
Enter the number of kg of U present:
kg U = 140 from above. This is estimated only.
Enter the number of kg of Ca remaining:
Since Ca is the LR, all of it will be consumed and none will be remaining.
Enter the number of kg of CaF2 present:
Convert mols Ca to mols CaF2. That's estimated 1180 mols Ca x (2 mols CaF2/2 mols Ca) = estimated 1180 mol CaF2.
kg CaF2 present = 1180 mols x molar mass CaF2 x (1 kg/1000g) = ?
Enter the number of kg of UF4 remaining:
Use the coefficients to convert mols Ca used to mols UF4 used.
1180 mols Ca x (1 mol UF4/2 mol Ca) = estimated 590 mols UF4 used.
mols UF4 initially = 930 mols.
mols UF4 used = 590 mols.
mols UF4 remaining = 930-590 = estimated 340 mols.
kg UF4 remaining = estimated 340 x molar mass UF4 x (1 kg/1000g) = ?
To answer this question, we need to determine the stoichiometry of the reaction, which tells us the ratio of reactants and products in the balanced chemical equation. From there, we can calculate the quantities of the reactants and products at the completion of the reaction.
The balanced chemical equation for the reaction is:
UF4 + 2Ca → U + 2CaF2
This equation tells us that 1 mole of UF4 reacts with 2 moles of Ca to produce 1 mole of U and 2 moles of CaF2.
First, let's convert the masses of the reactants to moles using their molar masses.
Molar mass of UF4:
(U: 238.03 g/mol) + (4 * (F: 18.998 g/mol)) = 352.03 g/mol
Molar mass of Ca:
40.08 g/mol
Number of moles of UF4:
292.0 kg * (1000 g/kg) / 352.03 g/mol = 829.68 mol
Number of moles of Ca:
47.0 kg * (1000 g/kg) / 40.08 g/mol = 1170.66 mol
Now, we can determine the limiting reactant, which is the reactant that is completely consumed in the reaction. The reactant that produces fewer moles of product is the limiting reactant.
From the balanced equation, we can see that 1 mole of UF4 produces 1 mole of U, while 1 mole of Ca produces 1 mole of U as well. Therefore, the limiting reactant is UF4.
Since the reaction goes to completion, all of the UF4 will be consumed. Thus, the number of moles of U produced is equal to the number of moles of UF4 initially present:
Number of moles of U: 829.68 mol
Next, using the stoichiometry of the reaction, we can calculate the number of moles of CaF2 produced:
Number of moles of CaF2: 2 * 829.68 mol = 1659.36 mol
Now, let's convert the moles of U and CaF2 back to masses:
Mass of U:
829.68 mol * (238.03 g/mol) = 197,250.8 g = 197.25 kg
Mass of CaF2:
1659.36 mol * (40.08 g/mol + 2 * (18.998 g/mol)) = 119,011.9 g = 119.01 kg
Finally, to calculate the remaining mass of Ca, we subtract the mass of U produced from the initial mass of Ca:
Remaining mass of Ca: 47.0 kg - 197.25 kg = -150.25 kg (negative value indicates that Ca is completely consumed)
Therefore, the reactor contents at the completion of this reaction are as follows:
- The number of kg of UF4 remaining: 0 kg
- The number of kg of Ca remaining: -150.25 kg
- The number of kg of U present: 197.25 kg
- The number of kg of CaF2 present: 119.01 kg